Question

What is the derivative of \[f(x)=\ln \left( {{x}^{2}}+3x-5 \right)\].

Answer 1

Hint: For solving this question we have to know the proper application of chain rule and some basic derivative formulas. We have to know that \[\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a\]. where \[a\] is a variable. Also, we should avoid calculation mistakes to get accurate answers.

Complete step by step solution:
From the question we were given that
\[\Rightarrow \]\[f(x)=\ln \left( {{x}^{2}}+3x-5 \right)\]…………..(1)
From the question it is clear that we have find derivative of \[\ln \left( {{x}^{2}}+3x-5 \right)\]
Now let us assume \[f\left( x \right)=y\]
So the equation (1) becomes \[y=\ln \left( {{x}^{2}}+3x-5 \right)\]……………(2)
So, now we have to find \[\dfrac{dy}{dx}\].
Take the equation \[y=\ln \left( {{x}^{2}}+3x-5 \right)\]
Now let us differentiate on both sides,
So, we will get equation as
\[\Rightarrow \]\[\dfrac{d}{dx}y=\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)\]……………..(3)
From the basic derivative formulas, we know that \[\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a\]. where \[a\]is variable
So now let us try to apply this formula for solving equation (3)
So, We will get
\[\Rightarrow \]\[\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{1}{{{x}^{2}}+3x-5}\times \dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)\]………….(4)
Now we will apply chain rule to solve. \[\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)\]
From the chain rule we know that
\[\Rightarrow \]\[\dfrac{d}{dx}\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+............+{{a}_{n}} \right)=\dfrac{d}{dx}{{a}_{1}}+\dfrac{d}{dx}{{a}_{2}}+\dfrac{d}{dx}{{a}_{3}}+............+\dfrac{d}{dx}{{a}_{n}}\]
Now we can write
\[\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=\dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}3x-\dfrac{d}{dx}5\]…………….(5)
We know that derivative of constant is ZERO, so \[\dfrac{d}{dx}5=0\].
We know that \[\dfrac{d}{dx}{{a}^{2}}=2a\times \dfrac{da}{dx}\]. where \[a\]is variable
 So, we can write \[\dfrac{d}{dx}{{x}^{2}}=2x\times \dfrac{dx}{dx}\] .as \[\dfrac{dx}{dx}=1\] it can be modified as \[\dfrac{d}{dx}{{x}^{2}}=2x\]
so, \[\dfrac{d}{dx}{{x}^{2}}=2x\].
Also, we know that \[\dfrac{d}{dx}ka=k\times \dfrac{d}{dx}a\],where \[a\]is variable and \[k\]=constant
So, we can write \[\dfrac{d}{dx}3x=3\times 1\] as \[\left[ \dfrac{dx}{dx}=1 \right]\]
So (5) can be written as \[\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=2x+3-0\]
\[\Rightarrow \]\[\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=2x+3\]……………………(6)
Put (6) in (4) we get
\[\Rightarrow \] \[\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{1}{{{x}^{2}}+3x-5}\times \left( 2x+3 \right)\]
\[\Rightarrow \]\[\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{2x+3}{{{x}^{2}}+3x-5}\]
So, finally we can conclude that the derivative of \[\ln \left( {{x}^{2}}+3x-5 \right)\] is \[\dfrac{2x+3}{{{x}^{2}}+3x-5}\]
\[\Rightarrow \]\[\dfrac{d}{dx}f(x)\]= \[\dfrac{2x+3}{{{x}^{2}}+3x-5}\]

Note: While solving this type of questions students should use correct formulas. using wrong formulas can lead to wrong answers. Students may have misconception that \[\dfrac{d}{dx}\ln a=a\left( \ln a-1 \right)\dfrac{d}{dx}a\], which is formula for integration of \[\ln x\] which is completely WRONG concept and the RIGHT formula is \[\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a\]